A cell has emf 1.5 V and internal resistance 0.25 Ω. What current flows if its terminals are joined by a wire of negligible resistance?
Correct answer A. 6.0 A
Explanation
Short-circuit current I = E/r = 1.5 V ÷ 0.25 Ω = 6.0 A. Multiplying instead of dividing gives 0.38 A, a common slip.
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